Given $\Bbb P(A\cap B') $ and $ \Bbb P(A'|B')$, how to find $\Bbb P(B)$?

Question

A and B are both events. Given $\Bbb P(A\cap B') = 1/6 $ and $ \Bbb P(A'|B') = 1/4$, what is $\Bbb P(B)$?

I've been trying all sort of conversions, but I still can't figure out how to solve this because of the complements...

I'm pretty sure the following formula will be useful, but that's it.

$ \Bbb P(B) = \Bbb P(A \cap B)/ \Bbb P(A|B)$

Thanks in advance.

Edit: Fixed typo

Answer 1

Since $P(A|B')+P(A'|B')=1$ we have $$P(A|B')=\dfrac{3}{4}$$also$$P(A|B')=\dfrac{P(A\cap B')}{P(B')}=\dfrac{P(A\cap B')}{1-P(B)}=\dfrac{\dfrac{1}{6}}{1-P(B)}=\dfrac{3}{4}\to\\P(B)=\dfrac{7}{9}$$